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DISCUSSION.  It is evident that both v and cb are constant. Therefore the motions of both the cylinder and the falling body are uniformly accelerated, the one in rotation, the other in translation.
When m is negligible compared with M, v is very small and consequently the motion very slow. When m is very large compared with M, v is practically equal to g, hence the body falls almost freely.
The linear acceleration of the axis of the cylinder equals one-half that of the falling body. The linear accelerations of the cylinder and of the falling body depend upon the radius of the cylinder only indirectly, i.e., through the mass of the cylinder.
4. A circular hoop is projected along a rough horizontal plane with a linear velocity VQ and an angular velocity oj0. Discuss the motion.
The hoop is acted upon by two forces, namely, its weight and the reaction of the plane. The latter may be resolved, as usual, into its normal component N and its frictional component F. Then the force equation gives
m|=F                               (1)
for the horizontal direction and
0 = N - mg                                         (2)
for the vertical direction.   On the other hand the torque equation gives
I.$-*Fa,                            (3)
where a is the radius of the hoop and Ic its moment of inertia about its own axis. The double sign indicates the fact that the direction of F changes with the direction in which slipping takes place at the point of contact. Denoting the coefficient of friction at the point of contact by & we have
F = vN, =               [by equation (2)].
Making this substitution hi equations (1) and (3) and replacing 7C in equation (3) by its value we obtain
and                                    "T. = T - g.                                             (5) 
Case /.  Suppose the initial angular velocity to be clockwise and